Multiplying Polynomials - Online Tutor, Practice Problems & Exam Prep

Multiplying polynomials involves methods like the distributive property and the FOIL method for binomials. The FOIL acronym stands for First, Outer, Inner, and Last, guiding the multiplication of two binomials. For polynomials with more than two terms, distribute each term of the shorter polynomial across the longer one. Special product formulas, such as the difference of squares and squares of binomials, simplify multiplication by providing quick solutions. Recognizing these patterns can save time and effort in calculations.

FOIL

Welcome back, everyone. We already saw how to add and subtract polynomials. And in this video, we're going to start talking about multiplying polynomials. It turns out we've already done some of this stuff before when we looked at the distributive property with expressions and exponents. We saw how to take terms on the outside of parentheses and distribute either constants or variables on the inside, but a lot of problems are not going to look like that. A lot of problems are actually going to start looking like this where you have 2 parentheses that are multiplied together. So a lot of problems are actually going to involve multiplying 2 binomials. Remember, these are just expressions with 2 terms. And so when we have these kinds of problems, we're not going to use the distributive property. We're going to use a new method called the FOIL method, and that's a term you may have heard in a math class somewhere. We're going to go over it, and I'm going to show you that it's actually a very straightforward procedure that tells you exactly what to do and the order. Alright. So let's just go ahead and jump right in.

FOIL is actually just an acronym, and it tells you which two terms to multiply with this expression over here, these 2, these 2 binomials, and it also tells you what order to multiply them in. Alright? So the F O I L, they stand for different things. Let's just jump right into it. The F actually stands for the first terms. So you multiply the first two terms of each expression. Some of the words that would be just the x and the x. The next thing you would do is the O, which is the outer terms. So if you think about this, there's sort of 4 terms in this expression or this multiplication. The outer terms are the 2 terms sort of on the outsides. So the x and the 3. Alright? Now the third thing is you're going to multiply the I, which is the inner terms. That would actually just be the 2 and the x that's on the inside right here. And the very last thing is you multiply the last terms. So you just multiply the 2 and the 3. Notice how every single term gets multiplied by the two terms in the opposite expression, so everything actually gets multiplied by each other. And that's how you sort of do it.

So let's just go ahead and FOIL this expression out. How do you multiply x plus 2 and x plus 3? Well, remember, F means that we're going to multiply the first two terms, the x and the x. What happens when you get that? You just get x². So let's do the O, which is the outer terms. That would just be the x and the 3. Well, x and 3 multiplied together just becomes 3x. What about the inner terms, the 2 and the x? That's the I. So the 2 and the x just becomes 2x. And finally, the 2 and the 3, those are the last terms. 2 multiplied by 3 just becomes 6, not 6x because they're just two numbers together. Alright? So that's it. That's the whole thing. That's how to FOIL.

The very last thing you have to do is once you FOIL, you're usually going to have to simplify your expression because usually what happens is these middle terms will be like terms, and you'll be able to combine them. Alright? So, in other words, this expression simplifies down to x². I can't combine that with anything else, but I can combine the 3x and the 2x because they're like terms, and they actually just combine down to 5x. And then finally, you just have the 6 that comes down like this, and that is your simplified expression. So this is how you FOIL 2 binomials and simplify it. Let me know if you have any questions. If not, let's move on to the next video.

Multiply the polynomials using FOIL.

$\left(x-5\right)\left(x-12\right)$

• A. $x^2-12x+60$
• B. $x^2-17x+60$
• C. $x^2-5x+60$
• D. $x^2-17x-60$

Multiply the polynomials using FOIL.

$\left(4x+7\right)\left(-x+6\right)$

• A. $4x^2+17x+42$
• B. $4x^2+31x+42$
• C. $-4x^2+17x+42$
• D. $-4x^2-x+42$

Multiply the polynomials using FOIL.

$\left(x^2-3x\right)\left(2x+8\right)$

• A. $2x^2+2x-24$
• B. $2x^3+2x^2-6x-24$
• C. $2x^3-2x^2-24x$
• D. $2x^3+2x^2-24x$

Multiply Polynomials Using the Distributive Property

Welcome back, everyone. In this video, I'm going to give you a summary of all the different ways that we multiply polynomials. For example, we saw how to distribute one term that's outside of parentheses by making sure that it multiplies into everything on the inside. We also saw in the last few videos how to foil something. That's where you have a 2 term times a 2 term expression. But some expressions aren't going to look like that. In fact, sometimes, you might have many terms like this 2 term expression multiplied by a 3 term expression. I'm going to show you how to work through this, and, basically, we're going to see that it's very familiar. Let's check it out.

So why can't I take an expression like this and foil it? Well, if you try to do this, what's going to happen is you'll do first and you'll do outer, then you'll do inner and then last. The problem is that once you've multiplied everything, this term over here on the inside actually never got multiplied by everything. You've kind of left it out. So foiling only actually works very specifically for 2 term times 2 term expressions. So that's the whole thing. Can't foil here. If you try to foil it, you're going to get the wrong answer.

So how do I deal with something like this? Well, basically, the idea is that when you multiply polynomials with more than 2 terms, all you really are going to do is you're going to split the terms of the shortest expression, and you're going to turn it into a distributive property type problem. So, sorry, then distribute. So here's what I mean by this. We're going to take this smaller expression like x+3, and I'm basically just going to break it up into 2 things. I'm gonna have an x over here and a 3 over here, and I'm basically just going to break it up into 2 distributive property type problems. So what goes on the inside here, it's this whole entire expression, x2+x-2 and x2+x-2. Alright? So, basically, what I'm doing here is I'm making sure that this x gets multiplied by everything that's on the inside. That's what happens in this first expression, And then the second one, making sure that this 3 multiplies everything that's on the inside, and that's what the second expression does. Alright? So, basically, what we've done here is we've turned this problem back into just one of the distributive property problems. That's the whole thing. The rest really is just doing the multiplication and stuff like that. So let's actually go ahead and do it.

Well, actually, for this one, I don't have to because notice how this is the exact same expression that I have over here. So this is what that becomes. I'm just going to rewrite this here, x3 + x2 - 2x. And what about this 3? Now we just have to distribute the 3 into the x2, that becomes 3 x2, Then the 3 into the x becomes 3x, and then the 3 into the negative 2 becomes minus 6. Alright? So now the rest is really just simplifying, but let's go ahead and do that. So I see an x3 term over here, but then I see an x or sorry, x3, and I have an x squared with an x squared, and I see a 2 negative two xs and a three x, and then I see a negative 6 off by itself. So remember, I'm just going to combine like terms. I have to combine things that are similar, and what this becomes here is this becomes x3, and then the x and the 4x2 becomes 4x2. The negative 2x and 3x become just x, and then I have negative 6 over here. So this is my simplified expression, and that's it. That's how you multiply polynomials just in general. And that's really all there is to it. So I'm actually just going to point out a pattern that's happened here, which has to do with the number of terms that we initially started with and ended with. So, in other words, when I took something like this, like a term outside and multiplied it into a 3 term expression, I did 1 times 3, and what I ended up getting here was I end up getting 3 terms, as my answer. And when I foiled, I did something similar. So I took a 2 term and a 2 term problem, and when I did the multiplication out, I ended up with 4 terms over here before I simplified it. The exact same thing happened over here. I took a 2 term expression, multiplied by a 3 term expression, and I end up just getting 6 terms before I did the simplification. So it's a really good way to check really easily when you've done the multiplication correctly. Always just take the number of terms that you have, multiply them, and whatever number you get, that's how many terms that you should get before you start simplifying everything down. If you don't have that, that means you dropped some, you've missed 1, or you may have counted an extra one.

 Multiply the polynomials.

$\left(x+4\right)\left(3x^2-2x+1\right)$

• A. $3x^3+10x^2-7x+4$
• B. $3x^3+12x^2+x+4$
• C. $12x^2-8x+4$
• D. $3x^3-2x^2+x$

Multiply the polynomials.

$\left(x+3\right)\left(x-5\right)\left(-2x+1\right)$

• A. $x^2-2x-15$
• B. $-2x^3+4x^2+30x$
• C. $-2x^3+5x^2+28x-15$
• D. $2x^3+5x^2+28x-15$

Special Products - Square Formulas

Welcome back, everyone. So by now, we've seen how to multiply polynomials by either foiling or distributing. What I'm going to show you in this video is that some multiplication problems will actually fit a special pattern, and when this happens, we can use formulas, these things called special product formulas, to multiply because it's going to be way easier and way faster than having to do stuff the long and tedious way by foiling or distributing. I'm going to show you that you can take a look at a problem like this, and very quickly, you'll be able to tell me that the answer is x2-25. I'm going to show you how it works. So, basically, if I take a look at an expression like this, the only way I know how to multiply it right now is by foiling. That's what I showed you. That's the, first and outer and then do the inner and last. So I'm going to multiply all these terms out to show you. This just becomes x2-3x+3x-9. Alright? And if you simplify this, what you'll see is that the negative 3x and positive 3x cancel, and all you end up with is x2-9. So that was a lot of work just to figure out that 2 of the middle terms will cancel, and the answer is relatively straightforward. It's just 2 terms. Now what actually happens is the reason this happened is because these two terms are the same, but just opposite sides. And whenever that happens, the middle terms are always actually going to cancel. So this is like a little bit of a pattern. This always happens as long as these two numbers are the same, and so we've come up with actually a special formula for this. And it's basically if you can sort of see if you can detect that a multiplication problem fits fits the sort of pattern of a+b times a-b, the answer will always just be a2-b2. This equation pops up a lot in math. We actually give it a name. It's called the difference of squares, but you don't need to know the name. So, basically, the whole thing is you can take a look at a problem, and if you can figure out that it fits this pattern, then you could just go straight to this rather than having to foil it, and that's going to be your answer. So let's take a look here. I have x+5 times x-5. So in other words, I have the same number, just opposite signs, so this is like a+b, a-b. So the answer is is just going to be a2-b2. So, basically, what happened is that, like, a is my x and b is my 5, and so, therefore, what happens is when I take a2, that just becomes x2, so these two things are the same, and when I do b2, that just becomes 25. Alright? And this is just your answer. Alright? So, basically, the way that these problems work out is it's kind of like pattern matching. Anytime you see a multiplication problem and you see the same numbers but opposite signs or you see stuff that's squared, usually, it's going to fit one of these formulas and you could just go straight here instead of having to multiply everything out the long way. Let's take a look at this next problem here, x+6 squared. Alright. This actually fits a special pattern, of a+b squared or ab-b squared. These equations are called the squares of binomials, and I'm just going to show you how these work out. This is basically what the right side equals, and we have a plus and plus as the two signs and a minus and a plus. One easy way to remember this, by the way, is that the first sign will always be the same as the left side, and the second sign will always be positive. So that's one way you can remember this real quick. Alright. So if I take x+6 squared, which one is this? Which one of the two patterns is this? Well, it's clearly this one over here, so I'm just going to basically use my formula. So this is a, and this is b. So in other words, my a is x. My b is 6. So what the answer is supposed to be is the answer is supposed to be a2+2ab+b2. All I have to do is now just, you know, replace these letters here with, with their x's and sixes and stuff like that. So my a2 just becomes x2. What about this 2 ab? Well, this just becomes 2. Then what's a? a is x, And then what's b? b is 6, and then + b2 ends up being 36. So if you simplify this, what you end up getting is x2+12x+36, and that is how to multiply this, again, very quickly without having to foil this. Alright, folks. So that really, that's all there is to it.

Multiply the polynomials using special product formulas.

$\left(5x-9\right)\left(5x+9\right)$

• A. $5x^2-81$
• B. $25x^2-81$
• C. $25x^2+90x+81$
• D. $25x^2-90x+81$

Multiply the polynomials using special product formulas.

$\left(3x+5\right)\left(3x+5\right)$

• A. $9x^2+15x+25$
• B. $9x^2+25$
• C. $9x^2+30x+25$
• D. $9x^2+55$

Special Products - Cube Formulas

Welcome back, everyone. In this video, we're going to continue talking about special product formulas, and in this video, we're going to cover some formulas involving cubes of polynomials rather than just squares. But the idea is the same. We're just going to be using these special products rather than having to multiply out the long way. Let's check it out. So, basically, I'm just going to jump right into the formulas over here. We have, cubes of binomials, which is a3+b cubed or a3-b cubed. Remember, we talked about squares of binomials. Now we just have cubes of binomials, and there's just more terms. But, basically, the way that this works is that the signs go plus, plus, plus, and then for a negative, you're going to do negative, plus, and negative. So one way to remember this is that for plus, everything is positive, and for minus, basically, the signs alternate. It goes negative, positive, and then negative. Okay? So, alright.

So if I take a look at my example here, I have x-3 cubed. So which one is it? It's basically just the a-b cubed. That's sort of the pattern that it fits. So in other words, this is like my a and that's my b. So, in other words, x is a and 3 is b. So what does this work out to? Well, this works out to I'm just going to write this out. x3-3x2b+3ab2-b3 . The rest is just I have to figure out which each one of these things becomes now that I know that a is x and b is 3. So let's get started. What is a cubed? Well, a cubed just becomes x3. Now what's the middle terms? So what does the first or the second term become? This becomes 3, and then a squared. If a is x, then a squared is x2, and then b is just 3. And then what about the 3rd term? This becomes 3. A is x, and then b is 3, so b squared just becomes 9. And then, minus b squared, so that's 3 times 3 times 3, which is just 27. Alright. So notice how fast this was. If we had to foil it out, it would have been a nightmare because we ended up with a 3 term, and then you gotta multiply that again. It just would have been a disaster. So these formulas are, again, relatively speaking, pretty pretty quick and straightforward. Alright.

So the last thing we just have to do is now simplify this. So the x cubed now just goes straight down. And what does the second term become? I have a 3 times a 3 times an x squared, so this just becomes 9x squared. The middle the third term actually goes to 3 times 27 times just 1 power x, so this is 27x, and then I have minus 27 over here. And that is the simplest I can make this. I can't combine any more terms or anything like that. Alright? Alright. So this is my expression over here, and that's how you do the cubes of binomials. Now with these types of formulas, almost always or you're almost never gonna have to memorize these. Your professors may actually give you something like a formula sheet. But in case they don't, I'm gonna give you a good way to think about this, and sort of memorize them. We've already talked about the signs. If you actually notice what happens to the coefficients, the coefficients always go 1, 3, 3, and 1. Right? So I have a 1 a cubes, a 3, and then 3, and then a 1 b cubed over here. So that's how the coefficients always go. There's also a pattern with the powers of a and the powers of b. The powers of a decrease as you go from left to right. So we have an a cubed, then an a squared, then an a to the one power. And then if you have no a's over here, it's basically like you have a to the zero power. So in other words, the powers of a always go from 3 to 1 to 0, and for b, it actually goes the opposite. The powers of b actually increase. Notice how here we have a b to the zero power, then b to the 1, then b to the 2, and b to the 3. So this goes 0, 1, 2, and 3. So, basically, a goes down and b goes up, and the coefficients are always 1, 3, 3, and 1.

Multiply the polynomials using special product formulas.

$\left(2x+4\right)^3$

• A. $8x^3+16x^2+32x+64$
• B. $8x^3+24x^2+24x+64$
• C. $8x^3+48x^2+96x+64$
• D. $x^3+24x^2+96x+64$